Lesson Explainer: The Quantization of Electromagnetic Radiation Physics • 9th Grade

In this explainer, we will learn how to calculate the energy of a photon given its frequency or wavelength.

Recall that light can be modeled as a wave.

The diagram below shows a wave traveling to the right, which could represent a light wave.

One cycle of the wave is shown in green.

The wavelength of the wave is the distance between any two corresponding points on the wave.

The frequency of the wave is the number of cycles that pass a point in space—for example, the dashed line—each second.

The speed is how fast a point on the wave moves through space. For this wave, we can think of this as how fast the orange dot moves to the right.

Recall that the speed, frequency, and wavelength of a wave are related. If the speed is 𝑣, the frequency is 𝑓, and the wavelength is 𝜆, then 𝑣=𝑓𝜆.

In free space, light moves at a constant speed of 3.00×10 m/s. This constant is given the label 𝑐, so for light 𝑐=𝑓𝜆.

The wave model of light is useful for describing phenomena such as refraction. However, there are some phenomena that this model cannot describe.

In the late nineteenth and early twentieth century, a phenomenon called the photoelectric effect was discovered. If a polished metal surface is exposed to light, electrons may be emitted from the surface of the metal. This must be done in a vacuum, otherwise the electrons will interact with and be absorbed by the air before they can be detected.

The diagram below shows an experimental setup that can be used to demonstrate the photoelectric effect.

A polished metal surface is inside a glass chamber. The air has been removed from the glass chamber, so that it is a vacuum. (It actually is not necessary to remove all the air from the chamber—the photoelectric effect can still be observed in a partial vacuum.)

The metal surface is part of a circuit containing a potential difference source and an ammeter. A wire connects the metal surface to the ammeter, another wire connects the ammeter to the potential difference source, and another wire connects the potential difference source to an anode in the glass chamber.

When the light source is off, there will be no electric current in the circuit, because the circuit is incomplete. There is a gap between the cathode—the metal surface—and the anode in the vacuum chamber. The potential difference source will create a potential difference across the cathode and anode, but electrons will not be able to flow between them.

When the light source is turned on, however, the light gives energy to the electrons in the metal surface, and they are able to leave the metal. Since electrons are negatively charged, and the anode is positively charged, they will be attracted to the anode and absorbed by it. Electrons can now flow between the cathode and anode. The circuit is complete, and there will be a current in it, which can be measured using the ammeter. This is the photoelectric effect.

The reading given by the ammeter for the current in the circuit is a measure of the number of electrons flowing past a point in the circuit each second, which is in turn a measure of the number of electrons being ejected from the metal surface each second. What factors in this experiment might we expect to influence the number of electrons being ejected each second and hence the reading on the ammeter?

We might expect the intensity of the light to affect the number of electrons ejected from the metal surface. Intensity is a measure of the energy transmitted by the light per unit time per unit area. Since electrons require energy to escape the metal surface, we might think that more energy per unit time per unit area of the metal surface would result in more electrons being ejected.

However, depending on the metal used for the metal surface—whether it is copper, aluminum, gold, or another metal—if we use a light source with a very low frequency, we will find that no electrons at all are ejected, no matter how bright we make the light source.

Conversely, if we use a light source with a very high frequency, we will find that some electrons will be emitted from the metal surface, even if the light source is very dim, and based on the intensity of the light source, we would expect it to take a very long time for enough energy to be absorbed by the metal surface.

So, for a given metal, if the light falling on the surface is below a certain frequency, the electrons in the metal will not be able to absorb the energy of the light to escape the metal.

This means that the energy of the light cannot be absorbed by the electrons continuously—otherwise an electron would have absorbed enough energy only after a certain amount of time has passed. An electron must have to absorb a certain amount of energy in one go in order to be ejected from the metal. Since this happens for higher frequency light, but not lower frequency light, the energy of the light must be delivered to the metal surface in discrete “packets,” and the amount of energy must be related to the frequency of the light.

These “packets” of energy are called photons. Photons are particles of light. We have found that the photoelectric effect must be described using a particle model of light, rather than a wave model.

Another name for photons is quanta. As the energy of light is divided into discrete quanta, we say that it is quantized.

The energy of a single photon is related to the frequency of the light. Specifically, if the frequency of the light is 𝑓, then the energy of a single photon, 𝐸, is given by 𝐸=ℎ𝑓.

Here, ℎ is a constant, known as the Planck constant. It is a constant of proportionality between the frequency of light and the energy of an individual photon. It has a very small value of 6.63×10 J⋅s. This means that a single photon is only a very small amount of energy, regardless of its frequency.

Formula: The Energy of a Photon given its Frequency

For light of frequency 𝑓, the energy, 𝐸, of a single photon is given by 𝐸=ℎ𝑓, where ℎ is the Planck constant, with a value of 6.63×10 J⋅s.

Since the frequency of light is related to its wavelength by 𝑐=𝑓𝜆, we can also express the energy of a photon in terms of the wavelength of the light. First, let’s rearrange 𝑐=𝑓𝜆 to make 𝑓 the subject as follows: 𝑓=𝑐𝜆.

Now, we can substitute this into our formula for the energy of a photon as follows: 𝐸=ℎ𝑐𝜆𝐸=ℎ𝑐𝜆.

Formula: The Energy of a Photon given its Wavelength

For light of wavelength 𝜆, the energy, 𝐸, of a single photon is given by 𝐸=ℎ𝑐𝜆, where ℎ is the Planck constant, with a value of 6.63×10 J⋅s, and 𝑐 is the speed of light in free space.

The energy of a photon is directly proportional to its frequency, so the higher the frequency, the greater the energy of the photon.

The energy of a photon is inversely proportional to its wavelength, so the longer the wavelength, the lower the energy of the photon. This means that, for example, a photon of red light has less energy than a photon of blue light, because red light has a longer wavelength than blue light.

The diagram below shows the spectrum of visible light. The energy of the light increases from the red end of the spectrum to the blue end.

Example 1: Finding the Energy of a Photon given its Frequency

What is the energy of a photon that has a frequency of 5.50×10 Hz? Use a value of 6.63×10 J⋅s for the Planck constant. Give your answer in scientific notation to two decimal places.

Answer

We can use the formula 𝐸=ℎ𝑓 to find the energy, 𝐸, of the photon, where 𝑓 is its frequency and ℎ is the Planck constant.

Substituting in the values given in the question, we get 𝐸=6.63×10⋅×5.50×10𝐸=3.6465×10⋅×.JsHzJsHz

Recall that Hzs=1, so JsHzJssJ⋅×=⋅×1=.

So, the energy of the photon is 3.6465×10 J, which when rounded to 2 decimal places is 3.65×10 J.

Example 2: Finding the Frequency of a Photon given its Energy

What is the frequency of a photon that has an energy of 2.52×10 J? Use a value of 6.63×10 J⋅s for the Planck constant. Give your answer in scientific notation to two decimal places.

Answer

The formula 𝐸=ℎ𝑓 relates the energy, 𝐸, of the photon, to 𝑓, its frequency, and ℎ the Planck constant.

First, let’s rearrange the equation to make 𝑓 the subject as follows: 𝑓=𝐸ℎ.

Now, we can substitute in the values given in the question as follows: 𝑓=2.52×106.63×10⋅𝑓=3.8009…×101.JJss

Recall that a unit of one over a second is equal to a unit of hertz. So, rounding to 2 decimal places, we get 𝑓=3.80×10.Hz

Example 3: Finding the Energy of a Photon given its Wavelength

What is the energy of a photon that has a wavelength of 400 nm? Use 6.63×10 J⋅s for the value of the Planck constant and 3.00×10 m/s for the value of the speed of light in free space. Give your answer in scientific notation to two decimal places.

Answer

We can use the formula 𝐸=ℎ𝑐𝜆 to find the energy, 𝐸, of the photon, where 𝜆 is its wavelength, ℎ is the Planck constant, and 𝑐 is the speed of light.

First, we need to convert the value given for the wavelength of the photon from nanometres to metres. Recall that 1=10nmm, so 400=400×10=4.00×10nmmm.

Substituting this value and the values given in the question into our formula, we get 𝐸=6.63×10⋅×3.00×10/4.00×10𝐸=4.9725×10⋅⋅/.JsmsmJsmsm

In the units, the factors of s and 1s in the numerator cancel, as does the factor of m in the numerator and denominator, leaving just J as follows: 𝐸=4.9725×10.J

Rounded to 2 decimal places, the energy of the photon is 4.97×10 J.

In a scenario where we have a large number of identical photons—that is, photons that all have the same frequency and wavelength—we can calculate the total energy of the photons simply by multiplying the energy of one photon by the number of photons.

If we have 𝑛 identical photons, then the total energy of the photons is given by 𝐸=𝑛ℎ𝑓 or 𝐸=𝑛ℎ𝑐𝜆.

Example 4: Finding the Total Energy of the Photons Produced by a Laser

A laser emits 4×10 photons, each with a frequency of 6×10 Hz. What is the total energy radiated by the laser? Use a value of 6.63×10 J⋅s for the Planck constant. Give your answer to the nearest joule.

Answer

We can use the formula 𝐸=𝑛ℎ𝑓 to find the total energy, 𝐸, of 𝑛 photons with frequency 𝑓, where ℎ is the Planck constant.

Substituting in the values given in the question, we get 𝐸=4×10×6.63×10⋅×6×10𝐸=159.12⋅⋅.JsHzJsHz

Recall that Hzs=1, so JsHzJssJ⋅⋅=⋅×1=.

The total energy of the photons produced by the laser is therefore 159 J to the nearest joule.

The discovery of the photoelectric effect was one of the first steps in the development of quantum mechanics.

Key Points

  • In order to describe certain physical phenomena, such as the photoelectric effect, light must be modeled as a particle.
  • Particles of light are called photons.
  • The energy of a photon is directly proportional to its frequency.
  • The energy of a photon is inversely proportional to its wavelength.
  • We can use 𝐸=ℎ𝑓 or 𝐸=ℎ𝑐𝜆 to find the energy of a single photon.
  • We can use 𝐸=𝑛ℎ𝑓 or 𝐸=𝑛ℎ𝑐𝜆 to find the energy of 𝑛 photons.

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